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2 years ago

What is the daughter nuclide when 0-15 experiences positron emission?

Chemistry

1 answer:

Helen [10]2 years ago

5 0

The daughter isotope (a decay product)of O-15 = N-15(Nitrogen 15)

<h3>Further explanation </h3>

Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,

alpha α particles ₂He⁴
beta β ₋₁e⁰ particles
gamma particles γ
positron particles ₁e⁰

O-15 emits positron particles ₁e⁰, so the atomic number decreases by 1, the mass number is the same

Reaction

$\tt _8^{15}O\Rightarrow _7^{15}N+_1^0e$

The mass number of the daughter isotope = 15, atomic number = 7

If we look at the periodic system, the element with atomic number 7 is Nitrogen (N)

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Alchen [17]

Answer:

T₂ = 84.375 K

Explanation:

Given data:

Initial volume = 3.3 L

Initial pressure = 2000 torr

Initial temperature = 225 K

Final temperature = ?

Final volume = 2.75 L

Final pressure = 900 torr

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂

T₂ = P₂V₂ T₁ /P₁V₁

T₂ = 900 torr× 2.75 L× 225 K / 2000 torr×3.3 L

T₂ = 556875 K/ 6600

T₂ = 84.375 K

7 0

2 years ago

An element has the electronic configuration 1s22s22p63s23p63d104s24p65s2. identify the element.

o-na [289]

The answer is Strontium

8 0

2 years ago

Problem PageQuestion Gaseous ethane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If of w

Nata [24]

Answer:

Explanation:

Balanced Equation: 2CH₃CH₃(g) + 5O₂(g) → 2CO₂(g) + 6H₂O(g)

Calculate moles of CH₃CH₃ and O₂

1.2 ₃₃ ( 1 ₃₃/ 30.0694 ₃₃ ) = 0.040 ₃₃

8.6 ₂ ( 1 2/ 31.998 ₂ ) = 0.27 ₃₃

Find limiting reagent 0.040 ₃₃ ( 5 ₂/ 2 ₃₃ ) = 0.10 ₂

CH₃CH₃ is the limiting Reagent

CH₃CH₃ (L.R.) O₂ CO₂ H₂O

Initial (mol) 0.040 0.27 0 0

Change (mol) -2x=-0 -5x= -0.10 +2x=+0.040 +6x=+0.12

Final (mol) 0 0.117 0.040 0.12

0.040 − 2 = 0 = 0.020

Determine percent yield

0.12 ₂ ( 18.0148 ₂ /1 ₂ ) = 2.2 ₂

= ( 1.08 / 2.2 ) 100% = 49%

5 0

3 years ago

calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?

inessss [21]

<u>Answer:</u>

$\Delta E=E_{final}-E_{initial}$

$\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}$

$\Delta E=-1312[0.111-1]KJ mol^{-1}$

$\Delta E=1166 KJ mol^{-1}$

$\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}$

$=193623 \times 10^{-23} \frac {J}{photon}$

$\Delta E=1.93623 \times 10^{-18} \frac {J}{photon}$

$\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}$

h is planck's constant

c is the speed of light

λ is the wavelength of light

$\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18} J/photon)}$

Wavelength

$\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m} =103 nm (Answer)$

<em>Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is </em><u><em>103 nm.</em></u>

7 0

2 years ago

1. Hydrogen and nitrogen combine to form ammonia. When nitrogen and hydrogen bond, nitrogen pulls the electrons from hydrogen to

UNO [17]

Answer:

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Explanation:

4 0

2 years ago